The elements of the first column are all multiples of $ x $; the items in the third column are all multiples of $ x-2 $. The determinant can be calculated in this way:

$$ 0 = left | begin {array} {ccc}

phantom {-} x & -2 and 3x-6 \

2x and ghost {-} 0 and 2-x \

-x and phantom {-} 5 and x-2

End {array} right | =

x (x-2) left | begin {array} {ccc}

phantom {-} 1 & -2 and phantom {-} 3 \

phantom {-} 2 and phantom {-} 0 & -1 \

-1 & phantom {-} 5 and phantom {-} 1

End {array} right | =

37x (x-2) quad implies quad x = 0.2 $$

To add some details …

The rules of the determinants allow to eliminate the common multipliers from a single column (or row):

$$ left | begin {array} {ccc}

pa & b & c \

pd & e & f \

ph & i & j

End {array} right | ; = ; p left | begin {array} {ccc}

a & b & c \

d & e & f \

h & i & j

End {array} right | $$

The justification is immediately clear if one is familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how the determining factors are expanded. Here is a quick example with $ 2 times 2 $:

$$ left | begin {array} {cc} pa & b \ pc & d end {array} right | = pa cdot d -pc cdot b = p (ad-cb) = p left | begin {array} {cc} a & b \ c & d end {array} right | $$

Each term in the expansion includes the multiplier, which can be decomposed.

However … For the problem in question, I calculated $ x $ and $ x-2 $ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it was not zero, in which case $ x $ could be *nothing*).

Source link

Neegggggggggggggg